Question

Show that the following point taken in order form the vertices of a rhombus.

(15, 20), (-3, 12), (-11, -6) and (7, 2)

Answer 1

Consider the given points.

$A(15,20),B(-3,12),C(-11,-6),D(7,2)$

So,

$AB=\sqrt{(-3-15{)}^2+(12-20{)}^2}$

$AB=\sqrt{(-18{)}^2+(-8{)}^2}$

$AB=\sqrt{324+64}=\sqrt{388}$

Similarly,

$BC=\sqrt{(-11+3{)}^2+(-6-12{)}^2}$

$BC=\sqrt{(-8{)}^2+(-18{)}^2}$

$BC=\sqrt{64+324}=\sqrt{388}$

Similarly,

$CD=\sqrt{(7+11{)}^2+(2+6{)}^2}$

$CD=\sqrt{(18{)}^2+(8{)}^2}$

$CD=\sqrt{324+64}=\sqrt{388}$

Similarly,

$DA=\sqrt{(15-7{)}^2+(20-2{)}^2}$

$DA=\sqrt{(8{)}^2+(18{)}^2}$

$DA=\sqrt{64+324}=\sqrt{388}$

Therefore,

$AB=BC=CD=AD$

Now,

$AC=\sqrt{(-11-15{)}^2+(-6-20{)}^2}$

$AC=\sqrt{(-26{)}^2+(-26{)}^2}$

$AC=\sqrt{576+576}=\sqrt{1152}$

Similarly,

$BD=\sqrt{(7+3{)}^2+(2-12{)}^2}$

$BD=\sqrt{(10{)}^2+(-10{)}^2}$

$BD=\sqrt{100+100}=\sqrt{200}$

So,

$AC\ne BD$

As all the sides are equal and diagonals are not equal.

Its shows that the following vertices are of rhombus.

Hence, this is the answer.